Answer:
[tex]E_x = \frac{2kQ}{\pi R^2}[/tex]
[tex]E_y = \frac{2kQ}{\pi R^2}[/tex]
Explanation:
Electric field due to small part of the circle is given as
[tex]dE = \frac{kdq}{R^2}[/tex]
here we know that
[tex]dq = \frac{Q}{\frac{\pi}{2}R} Rd\theta[/tex]
[tex]dq = \frac{2Q d\theta}{\pi}[/tex]
Now we will have two components of electric field given as
[tex]E_x = \int dE cos\theta[/tex]
[tex]E_x = \int \frac{kdq}{R^2} cos\theta[/tex]
[tex]E_x = \int \frac{k (2Qd\theta) cos\theta}{\pi R^2}[/tex]
[tex]E_x = \frac{2kQ}{\pi R^2} \int_0^{90} cos\theta d\theta[/tex]
[tex]E_x = \frac{2kQ}{\pi R^2} (sin 90 - sin 0)[/tex]
[tex]E_x = \frac{2kQ}{\pi R^2}[/tex]
similarly in Y direction we have
[tex]E_y = \int dE sin\theta[/tex]
[tex]E_y = \int \frac{kdq}{R^2} sin\theta[/tex]
[tex]E_y = \int \frac{k (2Qd\theta) sin\theta}{\pi R^2}[/tex]
[tex]E_y = \frac{2kQ}{\pi R^2} \int_0^{90} sin\theta d\theta[/tex]
[tex]E_y = \frac{2kQ}{\pi R^2} (-cos 90 + cos 0)[/tex]
[tex]E_y = \frac{2kQ}{\pi R^2}[/tex]
